Question

A first order reaction, where [a]0 = 1.00 m, is 74.7 % complete in 351 s. How long does it take for the same reaction to go from 1.00 m to 86.1 % completion?

Answer 1

ln ([A]t/[A]o) = -kt. . .where [A]t/[A]o is the fraction of reactant remaining at time t. If the reaction is 68.9% complete, then there is (100.0 - 68.9) = 31.1% of the original material remaining. 31.1% = 31.1/100 = 0.311. 

ln (0.311) = -(k)(339 s) 
-1.17 = -(k)(339 s) 
k = 0.00345 s^-1 

At 90.5% completion, 9.5% of the reactant remains. 

ln (0.095) = -(0.00345 s^-1)(t) 
-2.35 = -(0.00345)(t) 
t = 682 s

hope it will help you.....